If P = [ 3 2 1 2 - 1 2 3 2 ] and A = [ 1 1 0 1 ] and Q = P A P T and X = P T Q 2005 P , then X is equal to [2005]

Question

If $P = [\begin{matrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ - \frac{1}{2} & \frac{\sqrt{3}}{2} \end{matrix}]$ and $A = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}]$ and $Q = P A P^{T}$ and $X = P^{T} Q^{2005} P,$ then $X$ is equal to [2005]

Smart Achievers · Accepted Answer

(1)

Given, $P = [\begin{matrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ - \frac{1}{2} & \frac{\sqrt{3}}{2} \end{matrix}],$ $A = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}],$

$Q = P A P^{T}$ and $X = P^{T} Q^{2005} P$

Now, $Q = P A P^{T}$ $\Rightarrow Q^{2} = (P A P^{T}) (P A P^{T})$

$= P A (P^{T} P) A P^{T}$ $= P A (I A) P^{T}$ $= P A^{2} P^{T}$

Proceeding in the same way, $Q^{2005} = P A^{2005} P^{T}$

Now, $A = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}]$

Proceeding in the same way, $A^{2005} = [\begin{matrix} 1 & 2005 \\ 0 & 1 \end{matrix}]$

Now, $X = P^{T} Q^{2005} P = P^{T} (P A^{2005} P^{T}) P$

$= (P^{T} P) A^{2005} (P^{T} P)$ $= I A^{2005} I = A^{2005} = [\begin{matrix} 1 & 2005 \\ 0 & 1 \end{matrix}]$

Solution

Q.
If $P = [\begin{matrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ - \frac{1}{2} & \frac{\sqrt{3}}{2} \end{matrix}]$ and $A = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}]$ and $Q = P A P^{T}$ and $X = P^{T} Q^{2005} P,$ then $X$ is equal to [2005]

1 $[\begin{matrix} 1 & 2005 \\ 0 & 1 \end{matrix}]$

2 $[\begin{matrix} 4 + 2005 \sqrt{3} & 6015 \\ - 2005 & 4 - 2005 \sqrt{3} \end{matrix}]$

3 $\frac{1}{4} [\begin{matrix} 2 + \sqrt{3} & 1 \\ - 1 & 2 - \sqrt{3} \end{matrix}]$

4 $\frac{1}{4} [\begin{matrix} 2005 & 2 - \sqrt{3} \\ 2 + \sqrt{3} & 2005 \end{matrix}]$

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Solution

Q. If P=[3212-1232] and A=[1101] and Q=PAPT and X=PTQ2005P, then X is equal to [2005]

1 [1200501]

2 [4+200536015-20054-20053]

3 14[2+31-12-3]